Read the following paragraphs and answer the questions that follow: A capacitor is a system of two conductors separated by an insulator. In practice, the two conductors have charges \( Q \) and \( -Q \) with a potential difference \( V = V_1 - V_2 \) between them. The ratio \( \frac{Q}{V} \) is a constant, denoted by \( C \), and is called the capacitance of the capacitor. It is independent of \( Q \) or \( V \). It depends only on the geometrical configuration (shape, size, separation) of the two conductors and the medium separating the conductors. When a parallel plate capacitor is charged, the electric field \( E_0 \) is localized between the plates and is uniform throughout. When a slab of a dielectric is inserted between the charged plates (charge density \( \sigma \)), the dielectric is polarized by the field. Consequently, opposite charges appear on the faces of the slab, near the plates, with surface charge density of magnitude \( \sigma_p \). For a linear dielectric, \( \sigma_p \) is proportional to \( E_0 \). Introduction of a dielectric changes the electric field, and hence, the capacitance of a capacitor, and hence, the energy stored in the capacitor. Like resistors, capacitors can also be arranged in series or in parallel or in a combination of series and parallel.
Question: 1
Consider a capacitor of capacitance \( C \), with plate area \( A \) and plate separation \( d \), filled with air [Fig. (a)]. The distance between the plates is increased to \( 2d \) and one of the plates is shifted as shown in Fig. (b). The capacitance of the new system now is:
Show Hint
When the distance between the plates of a parallel plate capacitor is increased, the capacitance decreases. Additionally, if the effective area of overlap is reduced (as in this case where one plate is shifted), the capacitance is further reduced.
In this problem, we have a parallel plate capacitor with a capacitance \( C \) when the plates are separated by a distance \( d \). The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[
C = \frac{\epsilon_0 A}{d}
\]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between the plates.
When the distance between the plates is increased from \( d \) to \( 2d \), the capacitance will decrease because capacitance is inversely proportional to the distance between the plates. However, one of the plates is shifted, which changes the effective area of the capacitor. The shift reduces the effective area of overlap between the plates.
The capacitance can be expressed as:
\[
C' = \frac{\epsilon_0 A'}{2d}
\]
where \( A' \) is the effective overlapping area, which is now reduced. In this case, due to the plate shift, the effective overlapping area is halved. Thus, the new capacitance is given by:
\[
C' = \frac{C}{4}
\]
Therefore, the correct answer is \( \frac{C}{4} \).
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Question: 2
A slab (area A and thickness \( d_1 \)) of a linear dielectric of dielectric constant \( K \) is inserted between charged plates (charge density \( \sigma \)) of a parallel plate capacitor [plate area A and plate separation \( d>d_1 \)] and opposite charges with charge density of magnitude \( \sigma_p \) appear on the faces of the slab. The dielectric constant K is given by:
Show Hint
In problems involving dielectrics in capacitors, remember that the dielectric slab introduces an induced charge on its surfaces. The dielectric constant is determined by the ratio of the total charge densities, including the charges on the plates and the induced charges on the dielectric.
In this problem, we are dealing with a parallel plate capacitor with a dielectric slab inserted between the plates. The charge density on the plates is \( \sigma \), and the dielectric slab has an induced charge density \( \sigma_p \) on its faces. The dielectric constant \( K \) is a measure of the change in the electric field due to the dielectric material. To find \( K \), we need to consider the following:
1. Capacitor without Dielectric:
The electric field between the plates of the capacitor, when there is no dielectric, is given by:
\[
E_0 = \frac{\sigma}{\epsilon_0}
\]
where \( \sigma \) is the charge density on the plates and \( \epsilon_0 \) is the permittivity of free space.
2. Capacitor with Dielectric:
When a dielectric slab is inserted between the plates, the dielectric constant \( K \) modifies the electric field. The charge density on the plates is modified due to the presence of the dielectric, and the total effective charge density becomes \( \sigma + \sigma_p \), where \( \sigma_p \) is the induced charge density on the faces of the dielectric slab.
3. Dielectric Constant:
The dielectric constant \( K \) relates the electric field with the dielectric to the electric field without the dielectric. It is given by the ratio of the total charge densities, considering the induced charges on the dielectric. The formula for the dielectric constant \( K \) is:
\[
K = \frac{\sigma + \sigma_p}{\sigma}
\]
Thus, the dielectric constant is the ratio of the total charge density (sum of the charge density on the plates and the induced charge on the dielectric) to the original charge density on the plates.
Therefore, the correct expression for the dielectric constant \( K \) is \( \frac{\sigma + \sigma_p}{\sigma} \).
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Question: 3
An electric field \( E \) is established between the plates of an air-filled parallel plate capacitor, with charges \( Q \) and \( -Q \). \( V \) is the volume of the space enclosed between the plates. The energy stored in the capacitor is:
Show Hint
The energy stored in a capacitor can be calculated using the energy density in the electric field. The energy density is proportional to the square of the electric field. By multiplying the energy density by the volume of the space between the plates, we obtain the total energy stored.
To calculate the energy stored in a parallel plate capacitor, we can use the general formula for the energy stored in an electric field. The energy density \( u \) in the electric field is given by:
\[
u = \frac{1}{2} \epsilon_0 E^2
\]
where \( \epsilon_0 \) is the permittivity of free space and \( E \) is the electric field between the plates.
The total energy stored in the capacitor is the energy density multiplied by the volume \( V \) of the region between the plates. The volume of the space between the plates is given by \( V = A \times d \), where \( A \) is the area of the plates and \( d \) is the distance between the plates.
Thus, the total energy \( U \) stored in the capacitor is:
\[
U = u \times V = \frac{1}{2} \epsilon_0 E^2 \times V
\]
Hence, the correct answer is \( \frac{1}{2} \epsilon_0 E^2 V \).
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Question: 4
Three capacitors A, B, and M, each of capacitance \( C \), are connected to a capacitor N of capacitance \( 2C \) and a battery as shown in the figure. If the charges on A and N are \( Q \) and \( Q' \) respectively, then \( \frac{Q'}{Q} \) is:
Show Hint
A, B and M are in series (equivalent \(C/3\)) and N is across the battery. So \(Q = CV/3\) and \(Q' = 2CV\).
Step 1: Understanding the Question:
Four capacitors are drawn in the figure. A, B and M, each of capacitance \(C\), sit on the three sides of a loop. N, of capacitance \(2C\), is joined across the same two nodes as that loop, and the battery \(V\) is joined across both.
We need the ratio of the charge \(Q'\) on N to the charge \(Q\) on A.
Step 2: Key Formula or Approach:
For capacitors in series, \(\dfrac{1}{C_s} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots\), and every capacitor in the series has the same charge.
For a capacitor, \(q = CV\). Capacitors that are joined across the same battery get the same voltage.
Step 3: Detailed Explanation:
B, A and M lie one after the other between the two nodes, so they are in series:
\[ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \quad \Rightarrow \quad C_s = \frac{C}{3} \]
This chain is joined across the battery, so the charge on each of A, B and M is the charge of the chain:
\[ Q = C_s V = \frac{CV}{3} \]
Capacitor N is connected directly across the battery:
\[ Q' = (2C)V = 2CV \]
\[ \frac{Q'}{Q} = \frac{2CV}{CV/3} = 6 \]
Step 4: Why the other options are wrong:
Option (A) \(\frac{1}{6}\) is the upside-down ratio \(Q/Q'\).
Option (B) \(\frac{1}{3}\) comes from taking N across the chain as if it had the same voltage as A, or from forgetting the factor 2 in \(2C\).
Option (C) 3 comes from using \(Q' = CV\) for N, forgetting that its capacitance is \(2C\).
In none of these cases is the three-capacitor chain handled correctly. Only option (D) fits.
Final Answer:
The ratio of the charge on N to the charge on A is 6, which is option (D).
\[ \boxed{\frac{Q'}{Q} = 6} \]
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Question: 5
A slab (area A and thickness \( \frac{d}{2} \)) of dielectric constant \( K \) is inserted in a parallel plate capacitor of plate area \( A \) and plate separation \( d \). If \( C \) and \( C_0 \) are the capacitances of the capacitors with and without the dielectric, then \( \frac{C}{C_0} \) is:
Show Hint
When a dielectric slab is inserted into a parallel plate capacitor, the capacitance increases due to the dielectric constant. The total capacitance is the sum of the capacitances of the regions with and without the dielectric.
The capacitance \( C_0 \) of a parallel plate capacitor without any dielectric is given by the formula:
\[
C_0 = \frac{\epsilon_0 A}{d}
\]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.
When a dielectric slab of thickness \( \frac{d}{2} \) is inserted between the plates, the capacitance of the system changes. The dielectric constant of the material is \( K \), and the new capacitance \( C \) is affected by the dielectric.
The insertion of the dielectric divides the capacitor into two regions:
1. A region with the dielectric, which contributes a capacitance of:
\[
C_{\text{dielectric}} = \frac{K \epsilon_0 A}{\frac{d}{2}} = \frac{2K \epsilon_0 A}{d}
\]
2. A region without the dielectric, which has the same capacitance as the original system:
\[
C_{\text{no dielectric}} = \frac{\epsilon_0 A}{\frac{d}{2}} = \frac{2 \epsilon_0 A}{d}
\]
Thus, the total capacitance with the dielectric inserted, \( C \), is the sum of these two capacitances:
\[
C = C_{\text{dielectric}} + C_{\text{no dielectric}} = \frac{2K \epsilon_0 A}{d} + \frac{2 \epsilon_0 A}{d} = \frac{2 \epsilon_0 A}{d} (K + 1)
\]
Now, the ratio of the capacitances is:
\[
\frac{C}{C_0} = \frac{\frac{2 \epsilon_0 A}{d} (K + 1)}{\frac{\epsilon_0 A}{d}} = \frac{2(K + 1)}{1} = \frac{2K}{K + 1}
\]
Therefore, the correct answer is \( \frac{2K}{K + 1} \).
